Differentiate the function, equate to zero, and solve.
#f(x)=sinx#
#f'(x)=cosx#
#cosx=0 -> x=pi/2+npi# where #n# is an integer.
Now, the second derivative tells us that if #a# is a critical value of #f(x)# (a value which causes the derivative to go to zero), if #f''(a)>0,# there is a minimum at #(a,f(a)),# and if #f''(a)<0,# then there is a maximum at #(a, f(a))#.
As seen above, this function has infinitely many critical values due to the periodic nature of the function. But let's take #x=pi/2, (3pi)/2# and test them in #f''(x).#
#f''(x)=-sinx#
#f''(pi/2)=-sin(pi/2)=-1<0#
So, at #x=pi/2+2npi# there are maxima.
#f''((3pi)/2)=-sin((3pi)/2)=-(-1)=1>0#
So, at #x=(3pi)/2 + 2npi,# there are minima.
#f(pi/2)=1# so the local maxima are #(pi/2+2npi, 1)#
#f((3pi)/2)=-1# so there are local minima at #((3pi)/2+2npi, -1)#
